# Zero points of a solution

Define $y_\alpha (t)$ as a solution of the ODE $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 3y = 3$, with $y_\alpha (0) = \alpha$.

Now I am supposed to argue that if $y_\alpha (0) = \alpha < 0$, then there exist a $\beta >0$, with $y_\alpha (\beta) = 0$.

The general solution of the ODE is given by $y_{gen} (t) = c_1 * e^{-3t} + c_2 * e^{-t} + 1$.

Another question is wheter there are solutions $y(t)$ of the ODE with $t_2 > t_1 >0$, such that $y(t_1) = y(t_2) = 0$. So solutions with more than one zero point.

I get stuck with both questions. Any help is appreciated.

Well, it takes two constants to define a particular solution of a second-order ODE, so your $y_\alpha$ is in fact not a solution, but a whole bunch thereof. Not that it is really important, though.
To answer your first question, see what is the limit of any solution at $t\to\infty$.
To answer the second one... well, why won't you try and build some graphs for some arbitrary values of $c_1,c_2$? It might give you an idea where to look for the desired solution with two zeros.
• Since any solution is a lineair combination of the general solution, I guess that limit is equal to 1 if $t \rightarrow \infty$, because of the negative exponents. But what does that say about the existence of a $\beta >0$, with $y_\alpha (\beta) = 0$? – clubkli Oct 7 '15 at 18:25